# Invert actual bits of a number

Last Updated : 29 Mar, 2024

Given a non-negative integer n. The problem is to invert the bits of n and print the number obtained after inverting the bits. Note that the actual binary representation of the number is being considered for inverting the bits, no leading 0â€™s are being considered.

Examples:

`Input : 11Output : 4(11)10 = (1011)[2]After inverting the bits, we get:(0100)2 = (4)10.Input : 10Output : 5(10)10 = (1010)2.After reversing the bits we get:(0101)2 = (101)2        = (5)10.`

Method 1 (Using bitwise operators)
Prerequisite: Toggling k-th bit of a number

## C++

 `// CPP program to invert actual bits ` `// of a number. ` `#include ` `using` `namespace` `std; ` ` `  `void` `invertBits(``int` `num) ` `{ ` `    ``// calculating number of bits ` `    ``// in the number ` `    ``int` `x = log2(num) + 1; ` ` `  `    ``// Inverting the bits one by one ` `    ``for` `(``int` `i = 0; i < x; i++)  ` `       ``num = (num ^ (1 << i));  ` `  `  `    ``cout << num; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `num = 11;  ` `    ``invertBits(num); ` `    ``return` `0; ` `} `

## Java

 `// Java program to invert  ` `// actual bits of a number. ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` `    ``static` `void` `invertBits(``int` `num) ` `    ``{ ` `        ``// calculating number of  ` `        ``// bits in the number ` `        ``int` `x = (``int``)(Math.log(num) /  ` `                      ``Math.log(``2``)) + ``1``; ` `     `  `        ``// Inverting the  ` `        ``// bits one by one ` `        ``for` `(``int` `i = ``0``; i < x; i++)  ` `        ``num = (num ^ (``1` `<< i));  ` `     `  `        ``System.out.println(num); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `num = ``11``;  ` `        ``invertBits(num); ` `    ``} ` ` `  `} ` ` `  `// This code is contributed ` `// by Anuj_67 `

## Python3

 `# Python3 program to invert actual  ` `# bits of a number.  ` `import` `math ` ` `  `def` `invertBits(num):  ` ` `  `    ``# calculating number of bits  ` `    ``# in the number  ` `    ``x ``=` `int``(math.log2(num)) ``+` `1` ` `  `    ``# Inverting the bits one by one  ` `    ``for` `i ``in` `range``(x):  ` `        ``num ``=` `(num ^ (``1` `<< i))  ` `     `  `    ``print``(num)  ` ` `  `# Driver Code ` `num ``=` `11` `invertBits(num)  ` ` `  `# This code is contributed  ` `# by Rituraj Jain `

## C#

 `// C# program to invert  ` `// actual bits of a number. ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `void` `invertBits(``int` `num) ` `    ``{ ` `        ``// calculating number of  ` `        ``// bits in the number ` `        ``int` `x = (``int``)(Math.Log(num) /  ` `                       ``Math.Log(2)) + 1; ` `     `  `        ``// Inverting the  ` `        ``// bits one by one ` `        ``for` `(``int` `i = 0; i < x; i++)  ` `        ``num = (num ^ (1 << i));  ` `     `  `        ``Console.WriteLine(num); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `num = 11;  ` `        ``invertBits(num); ` `    ``} ` ` `  `} ` ` `  `// This code is contributed ` `// by Anuj_67 `

## Javascript

 ``

## PHP

 ` `

Output

```4

```

Time complexity : O(log n)
Auxiliary space : O(1)

Method 2 (Using bit-wise operators)

Think about generating a number just less than or equal to ‘n’ with all bits ‘1’ and take xor of that number with ‘n’.

Algorithm:

1. A variable ‘x’ is initialized to 1.
2. The function enters a while loop that continues until ‘x’ is less than or equal to the input integer (A).
3. Inside the while loop, ‘x’ is left-shifted by 1 using the bit-wise left shift operator (<<=). This multiplies ‘x’ by 2.
4. After the while loop, the value of ‘x’ is decremented by 1.
5. The function returns the result of the bit-wise XOR operation between ‘x’ and the input integer (A)
6. The final result is the bit-wise XOR of the input integer with the largest power of 2 that is less than or equal to the input integer.

Below is the implementation of the above idea:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `class` `GFG { ` `public``: ` `    ``int` `Opposite(``int` `n) ` `    ``{ ` `        ``int` `x = 1; ` `        ``while` `(x <= n) { ` `            ``x <<= 1; ` `        ``} ` `        ``x--; ` `        ``return` `(x ^ n); ` `    ``} ` `}; ` ` `  `int` `main() ` `{ ` `    ``int` `n = 5; ` `    ``GFG Obj = GFG(); ` `    ``cout << Obj.Opposite(n); ` `    ``return` `0; ` `}`

## Java

 `import` `java.util.*; ` `class` `GFG { ` `   ``public` `int` `Opposite(``int` `n) { ` `       ``int` `x = ``1``; ` `       ``while``(x <= n){ ` `           ``x <<= ``1``; ` `       ``} ` `       ``x--; ` `       ``return` `(x ^ n); ` `   ``} ` `} ` `public` `class` `Main { ` `   ``public` `static` `void` `main(String[] args) { ` `       ``int` `n = ``5``; ` `       ``GFG Obj = ``new` `GFG(); ` `       ``System.out.print(Obj.Opposite(n)); ` `   ``} ` `}`

## Python3

 `class` `GFG: ` `    ``def` `Opposite(``self``, n): ` `        ``x ``=` `1` `        ``while` `x <``=` `n: ` `            ``x <<``=` `1` `        ``x ``-``=` `1` `        ``return` `(x ^ n) ` ` `  `n ``=` `5` `Obj ``=` `GFG() ` `print``(Obj.Opposite(n)) `

## C#

 `using` `System; ` ` `  `public` `class` `GFG { ` `    ``public` `int` `Opposite(``int` `n) ` `    ``{ ` `        ``int` `x = 1; ` `        ``while` `(x <= n) { ` `            ``x <<= 1; ` `        ``} ` `        ``x--; ` `        ``return` `(x ^ n); ` `    ``} ` `} ` ` `  `class` `Program { ` `    ``static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int` `n = 5; ` `        ``GFG Obj = ``new` `GFG(); ` `        ``Console.WriteLine(Obj.Opposite(n)); ` `    ``} ` `} `

## Javascript

 `// Javascript code addition  ` ` `  `// Creating a class name GFG  ` `class GFG { ` `     `  `  ``Opposite(n) { ` `    ``let x = 1; ` `       `  `    ``// Keep multiplying x by 2, until x <= n ` `    ``while` `(x <= n) { ` `      ``x <<= 1; ` `    ``} ` `       `  `    ``// decrease the value of x by 1.  ` `    ``x--; ` `       `  `    ``// return xor of x and n.  ` `    ``return` `(x ^ n); ` `  ``} ` `} ` ` `  `// assigning value to n ` `let n = 5; ` ` `  `// creating an object.  ` `let Obj = ``new` `GFG(); ` ` `  `// calling Opposite() function from the class.  ` `console.log(Obj.Opposite(n)); ` ` `  `// The code is contributed by Arushi Goel.`

Output

```2

```

Time Complexity: O(log n)
Auxiliary Space: O(1)

Method 3 (Using Bitset)
Here we use the flip() of bitset to invert the bits of the number, in order to avoid flipping the leading zeroes in the binary representation of the number, we have calculated the number of bits in the binary representation and flipped only the actual bits of the number. We have used to_ulong() to convert bitset to number.

## C++

 `// CPP program to invert actual bits ` `// of a number. ` `#include ` `using` `namespace` `std; ` ` `  `void` `invertBits(``int` `num) ` `{ ` `    ``// calculating number of bits ` `    ``// in the number ` `    ``int` `x = log2(num) + 1; ` ` `  `    ``// Considering number to be 32 bit integer; ` `    ``bitset<32> b(num); ` ` `  `    ``// reversing the bits one by one ` `    ``for` `(``int` `i = 0; i < x; i++)  ` `        ``b.flip(i); ` ` `  `    ``// converting bitset to number ` `    ``cout << b.to_ulong();  ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `num = 11;  ` `    ``invertBits(num); ` `    ``return` `0; ` `} `

## Java

 `// Java program to invert actual  ` `// bits of a number. ` `class` `GFG  ` `{ ` `    ``static` `void` `invertBits(``int` `num) ` `    ``{ ` `        ``// calculating number of  ` `        ``// bits in the number ` `        ``int` `x = (``int``)(Math.log(num) /  ` `                      ``Math.log(``2``)) + ``1``; ` `     `  `        ``// Inverting the bits ` `        ``// one by one ` `         `  `        ``for` `(``int` `i = ``0``; i < x; i++)  ` `        ``num = (num ^ (``1` `<< i));  ` `     `  `        ``System.out.print(num); ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `num = ``11``;  ` `        ``invertBits(num); ` `    ``} ` `} ` ` `  `// This code is contributed by Mukul Singh `

## Python 3

 `# Python 3 program to invert actual  ` `# bits of a number. ` `import` `math ` ` `  `def` `invertBits(num): ` `     `  `    ``# calculating number of  ` `    ``# bits in the number ` `    ``x ``=` `int``(math.log(num, ``2.0``) ``+` `1``); ` ` `  `    ``# Inverting the bits ` `    ``# one by one ` `    ``for` `i ``in` `range``(``0``, x):  ` `        ``num ``=` `(num ^ (``1` `<< i));  ` ` `  `    ``print``(num); ` ` `  `# Driver code ` `num ``=` `11``;  ` `invertBits(num); ` ` `  `# This code is contributed  ` `# by Akanksha Rai `

## C#

 `// C# program to invert actual  ` `// bits of a number. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `    ``static` `void` `invertBits(``int` `num) ` `    ``{ ` `        ``// calculating number of  ` `        ``// bits in the number ` `        ``int` `x = (``int``)Math.Log(num, 2) + 1; ` `     `  `        ``// Inverting the bits ` `        ``// one by one ` `        ``for` `(``int` `i = 0; i < x; i++)  ` `        ``num = (num ^ (1 << i));  ` `     `  `        ``Console.Write(num); ` `    ``} ` `     `  `    ``// Driver code ` `    ``static` `void` `Main() ` `    ``{ ` `        ``int` `num = 11;  ` `        ``invertBits(num); ` `    ``} ` ` `  `} ` ` `  `// This code is contributed by Anuj_67 `

## Javascript

 ``

## PHP

 ` `

Output

```4

```

Time complexity: O(log n)
Auxiliary space: O(1)

Here, we will create a mask by setting all the bits from MSB(including MSB) and then take XOR with the original number

## C++

 `// CPP program to invert actual bits ` `// of a number. ` `#include ` `using` `namespace` `std; ` ` `  `void` `invertBits(``int` `num) ` `{ ` `    ``// calculating the mask ` `    ``int` `x = num;     ``// say num = 100000 ` `      ``x |= x >> 1;    ``// 100000 | 010000 = 110000 ` `      ``x |= x >> 2;    ``// 110000 | 001100 = 111100 ` `      ``x |= x >> 4;    ``// 111100 | 000011 = 111111 ` `      ``x |= x >> 8;    ``// 111111 | 000000 = 111111 ` `      ``x |= x >> 16;    ``// 111111 | 000000 = 111111 ` `  `  `    ``cout << (num ^ x); ``// 100000 | 111111 = 011111 ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `num = 11;  ` `    ``invertBits(num); ` `    ``return` `0; ` `} `

## Java

 `/*package whatever //do not write package name here */` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` `  ``public` `static` `void` `invertBits(``int` `num) { ` `    ``//base case ` `    ``if``(num == ``0``)  ` `    ``{ ` `      ``System.out.println(``1``); ` `      ``return``; ` `    ``} ` `    ``// calculating the mask ` `    ``int` `x = num;     ``// say num = 100000 ` `    ``x |= x >> ``1``;    ``// 100000 | 010000 = 110000 ` `    ``x |= x >> ``2``;    ``// 110000 | 001100 = 111100 ` `    ``x |= x >> ``4``;    ``// 111100 | 000011 = 111111 ` `    ``x |= x >> ``8``;    ``// 111111 | 000000 = 111111 ` `    ``x |= x >> ``16``;    ``// 111111 | 000000 = 111111 ` `     `  `    ``System.out.println(num ^ x); ``// 100000 | 111111 = 011111 ` `  ``} ` `  ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `num = ``11``;  ` `        ``invertBits(num); ` `    ``} ` `} ` ` `  `// This code is contributed by lapimpale@ `

## Python3

 `def` `invertBits(num): ` `    ``# calculating the mask ` `    ``x ``=` `num ` `    ``x |``=` `x >> ``1` `    ``x |``=` `x >> ``2` `    ``x |``=` `x >> ``4` `    ``x |``=` `x >> ``8` `    ``x |``=` `x >> ``16` ` `  `    ``print``(num ^ x) ` ` `  ` `  `# Driver Code ` `num ``=` `11` `invertBits(num) `

## C#

 `// C# code to implement the approach ` ` `  `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``public` `static` `void` `invertBits(``int` `num) ` `    ``{ ` `        ``// base case ` `        ``if` `(num == 0) { ` `            ``Console.WriteLine(1); ` `            ``return``; ` `        ``} ` `        ``// calculating the mask ` `        ``int` `x = num; ``// say num = 100000 ` `        ``x |= x >> 1; ``// 100000 | 010000 = 110000 ` `        ``x |= x >> 2; ``// 110000 | 001100 = 111100 ` `        ``x |= x >> 4; ``// 111100 | 000011 = 111111 ` `        ``x |= x >> 8; ``// 111111 | 000000 = 111111 ` `        ``x |= x >> 16; ``// 111111 | 000000 = 111111 ` ` `  `        ``Console.WriteLine(num ` `                          ``^ x); ``// 100000 | 111111 = 011111 ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int` `num = 11; ` `        ``invertBits(num); ` `    ``} ` `} ` ` `  `// This code is contributed by phasing17`

## Javascript

 `// JavaScript program to invert actual bits ` `// of a number. ` ` `  ` `  `function` `invertBits(num) ` `{ ` `    ``// calculating the mask ` `    ``let x = num;     ``// say num = 100000 ` `      ``x |= x >> 1;    ``// 100000 | 010000 = 110000 ` `      ``x |= x >> 2;    ``// 110000 | 001100 = 111100 ` `      ``x |= x >> 4;    ``// 111100 | 000011 = 111111 ` `      ``x |= x >> 8;    ``// 111111 | 000000 = 111111 ` `      ``x |= x >> 16;    ``// 111111 | 000000 = 111111 ` `  `  `    ``console.log(num ^ x); ``// 100000 | 111111 = 011111 ` `} ` ` `  `// Driver code ` `let num = 11;  ` `invertBits(num); ` ` `  ` `  `// This code is contributed by phasing17`

Output

```4

```

Time complexity: O(1)
Auxiliary space: O(1)

Method 5 (Extracting only the relevant bits using log and XOR)

The inverted number can be efficiently obtained by:

1. Getting the number of bits using log2

2. Taking XOR of the number and 2 numOfBits – 1

## C++

 `// CPP program to invert actual bits ` `// of a number. ` `#include ` `using` `namespace` `std; ` ` `  `void` `invertBits(``int` `num) ` `{ ` `   ``// Find number of bits in the given integer ` `   ``int` `numOfBits = (``int``)log2(num) + 1; ` ` `  `   ``// invert the number by taking  ` `   ``// xor of n and (2 raised to numOfBits) - 1 ` `   ``cout <<  (((1 << numOfBits) - 1) ^ num); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `num = 11;  ` `    ``//Function Call ` `    ``invertBits(num); ` `    ``return` `0; ` `} ` ` `  `//This code is contributed by phasing17 `

## Java

 `// Java program to invert actual bits ` `// of a number. ` `import` `java.io.*; ` ` `  `class` `GFG { ` `static` `void` `invertBits(``int` `num) ` `{ ` `// Find number of bits in the given integer ` `int` `numOfBits = (``int``)(Math.log(num) / Math.log(``2``)) + ``1``;  ` `      ``// invert the number by taking  ` `    ``// xor of n and (2 raised to numOfBits) - 1 ` `    ``System.out.println(((``1` `<< numOfBits) - ``1``) ^ num); ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `num = ``11``; ` `    ``//Function Call ` `    ``invertBits(num); ` `} ` `} `

## Python

 `# Python program to invert actual bits ` `# of a number. ` ` `  `def` `invertBits(num): ` `    ``# Find number of bits in the given integer ` `    ``numOfBits ``=` `num.bit_length() ` ` `  `    ``# Invert the number by taking ` `    ``# xor of n and (2 raised to numOfBits) - 1 ` `    ``print``(((``1` `<< numOfBits) ``-` `1``) ^ num) ` ` `  `# Driver code ` `num ``=` `11` `# Function Call ` `invertBits(num) `

## C#

 `// C# program to invert actual bits ` `// of a number. ` ` `  `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `void` `invertBits(``int` `num) ` `    ``{ ` `        ``// Find number of bits in the given integer ` `        ``int` `numOfBits = (``int``)(Math.Log(num) / Math.Log(2)) + 1; ` `        ``// invert the number by taking  ` `        ``// xor of n and (2 raised to numOfBits) - 1 ` `        ``Console.WriteLine(((1 << numOfBits) - 1) ^ num); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int` `num = 11; ` `        ``// Function Call ` `        ``invertBits(num); ` `    ``} ` `} ` ` `  ` `  `// contributed by phasing17`

## Javascript

 `// JavaScript program to invert actual bits ` `// of a number. ` ` `  `function` `invertBits(num) { ` `  ``// Find number of bits in the given integer ` `  ``let numOfBits = Math.floor(Math.log2(num)) + 1; ` `   `  `  ``// Invert the number by taking  ` `  ``// xor of n and (2 raised to numOfBits) - 1 ` `  ``console.log(((1 << numOfBits) - 1) ^ num); ` `} ` ` `  `// Driver code ` `let num = 11; ` `// Function Call ` `invertBits(num); `

Output

```4

```

Time complexity: O(1)
Auxiliary space: O(1)